Question

A 10 kg brick moves along an x axis. Its acceleration as a function of its position is shown in Fig.7-38. The scale of the figure’s vertical axis is set by as 20.0 m/s². What is the net work performed on the brick by the force causing the acceleration as the brick moves from $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$?

Short answer

The net work done on the brick by variable force $\mathrm{W}=800\mathrm{J}$

Step-by-step solution

Understanding the concept

We can use the equation of work done bythespring along with the equation for area under the curve for a graph.

Formula:

1. $\mathbf{W}\mathbf{=}\mathbf{F}\mathbf{\times }\mathbf{x}$
2. $\mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{base}\mathbf{\times }\mathbf{height}$

Given:

1. The mass of brick,$\mathrm{m}=10\mathrm{kg}$
2. The scale of acceleration is set as, ${\mathrm{a}}_{\mathrm{s}}=10\mathrm{m}/{\mathrm{s}}^2$

Calculate the work done

We have,

$$\begin{gathered} \mathrm{W}=\mathrm{F}.\mathrm{x} \\ \mathrm{W}=\mathrm{ma}.\mathrm{x} \\ \mathrm{W}=\mathrm{m}\left(\mathrm{a}\times \mathrm{x}\right) \end{gathered}$$

Here, the product $\left(\mathrm{a}\times \mathrm{x}\right)$is given from the graph, as the area under the curve of $\mathrm{a}\left(\mathrm{x}\right)$

So, $\mathrm{W}=\mathrm{m}\times \mathrm{A}$

Where

$\begin{array}{rcl}\mathrm{A}& =& \frac{1}{2}\times \mathrm{base}\times \mathrm{height}\\ & =& \frac{1}{2}\times 8.0\mathrm{m}\times 20.0\mathrm{m}\\ & =& 80{\mathrm{m}}^2\end{array}$

Hence,

$$\begin{gathered} \mathrm{W}=10\mathrm{kg}\times 80{\mathrm{m}}^2 \\ \mathrm{W}=800\mathrm{J} \end{gathered}$$

Therefore, the net work performed on the brick by the force 800 J.